Calculating Speed After Collision in Railroad Freight Yard

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The discussion focuses on calculating the speed of two coupled railroad freight cars after a collision. An empty freight car with mass m collides with a stationary loaded boxcar of mass 3.8m, initially moving at 1.0 m/s. The correct equation for the conservation of momentum is (m1 + m2) vf = m1v1. The final speed after the collision can be simplified to Vf = (m1)(1.0) / (m1 + 3.8), eliminating the variable m from the final expression.

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In the railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded boxcar of mass 3.8m. The two cars couple together upon collision.

(a) What is the speed of the two cars after the collision?



This is the equation I have devised so far:
m1= M
m2= 3.8m
V1= 1.0 m/s
p2= 0
pf=p1 or (m1 + m2 )vf = (m1v1)

Vf = (m1)(1.0) / (m1 + 3.8)

I don't think this is correct. With the first mass being indicate as only m, I'm not sure how to compensate. Could someone please let me know what I'm doing incorrectly?
 
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evan,

You're very close! Your equation:

(m1 + m2) vf = m1v1

is exactly right. Do you know what law of physics this equation comes from?

You're mistake is just in plugging in the given values for m1, m2. Try it again.
 
That "m" should not appear in the final expression.It should be simplified through.

Daniel.
 

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